# Thread: dollar and cent coins were mixed up. banker needs help!

1. ## dollar and cent coins were mixed up. banker needs help!

Dear sirs!
How much money has the client?
Can we help the banker?
This exercise is part of exercises about the subject: "Number theory" and it could be a diophantine equation.
A dispersed bank cashier mistook 1 dollar coins and 1 cent coins when he paid out the cheque of Mr. Brown, while he gave him 1 dollar coins instead of 1 cent coins and 1 cent coins instead of 1 dollar coins. After Mr. Brown had - at home - put generously 5 cents in the savings tin of his son, he discovered that he had exactly twice as much money left over, as had stood on the cheque. On which sum was the cheque drawn?
I've already got the following thoughts:
c= amount of 1 Cent coins
k= amount of 1 Dollar coins
2(100k+c) = k+100c-5
98c = 199k+5
c = 2.03061224k + 0.05263158
Now I tried the numbers from 1 to 500 for c, but k was never an integer number. But because there are only whole coins, and not ones like 0.05678, k and c have to be integer numbers.
Could you tell me what I have to correct?
Or is it just an impossible question?
Sincerly
enoimreh

2. Your only flaw here is that you've subjected yourself to rounding errors. 199/98 is not exactly equal to 2.03061224, and neither is 5/98 exactly equal to 0.05263158. As far as I can tell, there's no diophantine solutions to c = 2.03061224k + 0.05263158, but there is a solution (in fact there are infinitely many such solutions) to c = 199/98k + 5/98. If you try your previous brute force method again with the corrected equation, you'll see that c = 262, k = 129 is a solution.

The trick here is that we know 199/98k + 5/98 must be an integer. If we pull out the common factor of 1/98, we see that 1/98(199k + 5) must be an integer, and from that we can deduce that 199k + 5 must be a multiple of 98. Where do you think you'd go from here?

Edit: So I see there's perhaps been a bit of confusion about what I meant. Sorry about that. I agree with the other helpers that the number of each type of coin must be 99 or less (and obviously cannot be negative), rendering the solution I gave invalid. However, I didn't mean to imply that c = 262, k = 129 is a solution to the overall problem, only that it was a solution to that one equation, taken "in a vacuum". My hope was enoimreh7 would see why the solution I gave works with the equation with the exact fractional values but not with the decimal approximations, and from there would be able to solve the problem on their own.

3. Originally Posted by ksdhart2
Your only flaw here is that you've subjected yourself to rounding errors. 199/98 is not exactly equal to 2.03061224, and neither is 5/98 exactly equal to 0.05263158. As far as I can tell, there's no diophantine solutions to c = 2.03061224k + 0.05263158, but there is a solution (in fact there are infinitely many such solutions) to c = 199/98k + 5/98. If you try your previous brute force method again with the corrected equation, you'll see that c = 262, k = 129 is a solution.

The trick here is that we know 199/98k + 5/98 must be an integer. If we pull out the common factor of 1/98, we see that 1/98(199k + 5) must be an integer, and from that we can deduce that 199k + 5 must be a multiple of 98. Where do you think you'd go from here?
You're right, one must always use fractions, not decimal approximations, in solving a Diophantine equation. (enoimreh7, have you learned the standard method ksdhart2 is using here?)

It is also important to (a) consider other restrictions, in this case that c and k must be less than 100 as these problems are typically interpreted, so your solution is not really usable, and (b) realize that there are infinitely many solutions, and in fact another of them is the answer to the problem.

4. Originally Posted by Dr.Peterson
You're right, one must always use fractions, not decimal approximations, in solving a Diophantine equation. (enoimreh7, have you learned the standard method ksdhart2 is using here?)

It is also important to (a) consider other restrictions, in this case that c and k must be less than 100 as these problems are typically interpreted, so your solution is not really usable, and (b) realize that there are infinitely many solutions, and in fact another of them is the answer to the problem.
Hello! No, I didnt. Sincerly, Enoimreh7

5. Hello!
Yes, c, the amount of Cents has to be smaller than 100, but for the Dollars it doesnt really matter or does it(???). The banker is so confused. Nevertheless I think, the only practicable solution is:
c= 63
k=31
98*63=199*31+5
(?)
Thank you all very much for your help!
31 Dollar, 63 Cent
Sincerly, Enoimreh

6. Originally Posted by Dr.Peterson
It is also important to (a) consider other restrictions, in this case that c and k must be less than 100 as these problems are typically interpreted ...

Dear Mr. Peterson! Isnt it sufficient, to have the Cents smaller 100? Logically you cannot write more than 99 Cents on a cheque.
But more than 100 Dollar could be on a cheque and than the banker could hand out more than 100 Cents mistaking the amount of Dollars for Cents.
He might than realise the mistake, but ...?
Why do you say, both have to be smaller than 100?
Sincerly, Enoimreh

7. Originally Posted by enoimreh7
Dear Mr. Peterson! Isnt it sufficient, to have the Cents smaller 100? Logically you cannot write more than 99 Cents on a cheque.
But more than 100 Dollar could be on a cheque and than the banker could hand out more than 100 Cents mistaking the amount of Dollars for Cents.
He might than realise the mistake, but ...?
Why do you say, both have to be smaller than 100?
Sincerly, Enoimreh
Originally Posted by enoimreh7
A dispersed bank cashier mistook 1 dollar coins and 1 cent coins when he paid out the cheque of Mr. Brown, while he gave him 1 dollar coins instead of 1 cent coins and 1 cent coins instead of 1 dollar coins. After Mr. Brown had - at home - put generously 5 cents in the savings tin of his son, he discovered that he had exactly twice as much money left over, as had stood on the cheque. On which sum was the cheque drawn?
Possibly you are right for this version of the problem. Often it is expressed not as confusing the two kinds of coins, but as misreading (or even miswriting) the amount, exchanging dollars and cents in the number, as if he thought the amount was 63.31 instead of 31.63. If the actual number of cents were greater than 99, he would not be able to make that confusion.

That is what I was thinking when I wrote that. In your problem, he is just confusing the coins themselves, not the written numbers, so I think you are right.

(I didn't see this response when you sent it, because I was in the hospital and didn't visit the site until the next week. I just saw it now, when I looked at your history.)

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